Such was the number and variety of responses to this challenge that presenting a detailed breakdown of one such solution – as has been the case for all of the first eleven in this series of challenges – would, I feel, be somewhat inappropriate.

For the majority of these challenges, it could be argued that there has been one solution which is indisputably “better” than the rest. Perhaps such an adjudication can also be made here, though to do so would certainly not be a straightforward exercise. What’s more, to pick just one of the many solutions would be to leave the rest – unfairly in my opinion – left on the sidelines.

As such, I would refer the readers to the many solutions in that post and to enjoy dissecting the varied and wonderful constructions therein. And to simply thank all those – Alex, aMareis, Maxim, John Jairo, sam, Jeff, Lori, Ron, Michael, Christian and XLarium – whose excellent contributions led to such a fruitful and inspiring discussion.

There’s evidently still much to be discovered in the world of worksheet formulas!

Readers who have read some of my earlier posts will be familiar with the concept of “redimensioning” an array.

This is an extremely useful and important technique, which, in its basic form, allows us to take a two-dimensional array and convert it into one of just a single dimension, whilst of course retaining the elements within that array.

Such an approach is necessary if we wish to further manipulate the entries of some two-dimensional array. For example, we might be in a position in which, for whatever reason, we need to pass each of the entries in a two-dimensional array to an array of one or more parameters for further processing. However, since the evaluation of the resulting multi-dimensional “matrix” is not within Excel’s capabilities, we are obliged to first transform the original array to one of a single dimension.

Note to readers: this post has been updated due to the inclusion – at the request of Torstein – of a further version of this solution, in which the number of values to be considered is dynamic and so may be set by the user. This version may be found at the very end of this post.

This post, inspired by a question from Patrick MacKay, from Belgium – thanks, Patrick! 🙂 – is a (rather belated) follow-up to that which I made here, in which, to recap, I presented a formula-based set-up which, given a target figure plus a series of values, determined which, if any, combination of those values had a sum equal to the target.

The only slight drawback to that solution was the caveat that, if more than one combination of values existed which satisfied that condition, then only one of those combinations was given.

Here I would like to improve upon that set-up by presenting a refined version which will return all such combinations. What’s more, at the very end of this deconstruction I will give a further version of the solution in which the number of values to be considered is a variable which may be set by the user.

In fact, that early post was also one of the very few in which I did not give an explanation as to how the solution works, which I would like to do here.

As an example of the output, imagine that our target value – £1054.35, for example – is here in A1, and that we have a list of 10 values in A2:A11, as below:

One correct solution received, courtesy of Lori, who not only presented a fine construction for working in Excel 2010 and earlier, but also a 2013 version, which had the added benefit of taking advantage of some of the new (and evidently very useful) features of that version to noticeably abridge the required set-up. So many thanks to Lori for sharing this knowledge and also congratulations on an excellent solution to a particularly complex challenge!

This is a formula-based solution which, given a positive number in base 10, converts that number into its equivalent form for another, given positive base.

So, given a base-10 value to be converted in A2, and the base to which we wish to convert this value in B2, the required output is generated in C2. Cell D2 contains a “back-check” which re-converts the value in C2 to base 10.

This is a reasonably complex problem, and certainly so if we want to present a solution which is relatively concise. However, despite its complexity (and arguably lack of practical use), the solution demonstrates some important techniques for working with strings, and so is not without merit.

Sometimes we are in a situation where we have a target figure plus a series of values and we want to know which, if any, combination of those values has a sum which is equal to the target.

This can be done as follows:

Edit: this post has now been revised here to account for multiple returns, should that be a requirement.

Using the above set-up, with our target value in A2 and our (in this case 9) values in C1:K1, we will place formulas in C2:K2 which will contain an “X” if the value in the row above forms part of our solution.